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1. Question 1
A data processing pipeline consists of nnn services connected in a series where the output of the iiith service serves as the input to the (i+1)(i+1)(i+1)th service. However, the processing units have varying latencies, and the throughput of the iiith unit is represented by the array throughput[i]
in messages per minute.
The first service receives input through an API, and the nnnth service produces the final output.
Each service can be scaled up independently, with the cost of scaling up the iiith service one unit equal to scaling_cost[i]
. After scaling up a service xxx times, it can process throughput[i]*(1 + x)
messages per minute.
Given the arrays throughput
and scaling_cost
, both of size nnn, and an integer budget
representing the budget available, determine the optimal scaling configuration for the services such that the throughput generated by the nnnth service is maximized.
Example: For instance, throughput = [4, 2, 7]
, scaling_cost = [3, 5, 6]
, and budget = 32
.
To maximize the throughput of the final service, an optimal solution is:
Service Index | Scale From | Scale To | Times Scaled | Cost per Scaling | Total Cost |
---|---|---|---|---|---|
0 | 4 | 12 | 2 | 3 | 6 |
1 | 2 | 10 | 4 | 5 | 20 |
2 | 7 | 14 | 1 | 6 | 6 |
When these units are applied in series, they generate a throughput of 10 units, the maximum possible throughput given the budget. Hence the answer is 10.
Function Description:
Complete the function getMaximumThroughput
in the editor below.
getMaximumThroughput
has the following parameters:
int throughput[n]
: the throughput generated by each of the nnn servicesint scaling_cost[n]
: the cost of scaling up a service one timeint budget
: the available money
Constraints:
![](https://csoahelp.com/wp-content/uploads/2024/07/image-22-1024x314.png)
Input Format For Custom Testing:
Sample Input:
3
3 2 5
3
2 5 10
28
Sample Output:
6
Explanation:
Service Index | Throughput | Scale To | Times Scaled | Cost per Scaling | Total Cost |
---|---|---|---|---|---|
0 | 3 | 15 | 4 | 2 | 8 |
2 | 2 | 6 | 2 | 5 | 10 |
2 | 5 | 10 | 1 | 10 | 10 |
Hence the answer returned is [3, 2, 0, 0, 1].
2. Question 2
Implement a prototype of a friend recommendation system for a social media application.
There are nnn users indexed from 0 to n−1n-1n−1, and mmm friends represented as a 2D array, friendships
, where the iiith friendship is a connection between users friendship[i][0]
and friendship[i][1]
.
User xxx is suggested as a friend to user yyy if xxx and yyy are not friends and have the maximum number of common friends, i.e., a friend of both xxx and yyy. If there are multiple possible such users xxx, the one with the minimum index is suggested.
Given nnn and friendships
, for each of the nnn users, find the index of the friend that should be recommended to them. If there is no recommendation available, report -1
.
Example: Suppose n=5n = 5n=5, m=5m = 5m=5, and connections = [[0, 1], [0, 2], [1, 3], [2, 3], [3, 4]]
.
User | Max Common Friends With | Recommendation |
---|---|---|
0 | 3 (1, 2) | 3 |
1 | 2 (0, 3) | 2 |
2 | 1 (0, 3) | 1 |
3 | 0 (1, 2) | 0 |
4 | 2 (3, 1, 3) | 1 (minimum Index) |
第一道题涉及一个数据处理流水线优化的问题。我们需要在给定预算的情况下,通过扩展各个服务的处理能力,最大化流水线末端的处理吞吐量。推荐使用的算法方法包括:
- 动态规划:将问题分解成子问题,逐步求解每个子问题的最优解,并利用这些子问题的解构建出整个问题的最优解。
- 贪心算法:在每一步选择当前最优解,通过反复选择局部最优解来构建全局最优解,适用于某些特定情况下的优化问题。
第二道题是关于社交媒体的好友推荐系统。我们需要为每个用户找到一个推荐好友,这个好友与用户有最多的共同好友,并且两人之间没有直接的好友关系。推荐使用的算法方法包括:
- 图论算法:使用图数据结构表示用户及其好友关系,通过遍历图来查找共同好友。可以利用深度优先搜索(DFS)或广度优先搜索(BFS)来实现。
- 哈希表:使用哈希表存储和快速查找用户及其好友关系,有助于高效计算共同好友数。
These two algorithmic problems highlight the importance of efficient data manipulation and optimization in real-world applications like server scaling and cart management. By understanding and applying the right algorithms, we can ensure optimal performance and user experience.
If you have any questions or would like to discuss these problems further, contact us!
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